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An aperiodic set of 11 Wang tiles
Emmanuel Jeandel, Michael Rao
To cite this version:
Emmanuel Jeandel, Michael Rao. An aperiodic set of 11 Wang tiles. 2020. �hal-01166053v3�
An aperiodic set of 11 Wang tiles
Emmanuel Jeandel and Michaël Rao September 21, 2020
Abstract
We present a new aperiodic tileset containing 11 Wang tiles on 4 colors, and we show that this tileset is minimal, in the sense that no Wang set with either fewer than 11 tiles or fewer than 4 colors is aperiodic. This gives a definitive answer to the problem raised by Wang in 1961.
Wang tiles are square tiles with colored edges. A tiling of the plane by Wang tiles consists of placing a Wang tile in each cell of the gridZ2so that contiguous edges share the same color. The formalism of Wang tiles was introduced by Wang [Wan61] to study decision procedures for a specific fragment of logic (see Section1.1for details).
Wang posed the question of whether an aperiodic tileset exists: a set of Wang tiles which tiles the plane but cannot do so periodically. His student Berger quickly gave an example of such an aperiodic tileset, but it consisted of a very large number of tiles. The number of tiles needed was eventually reduced by Berger himself and then by others, to achieve an aperiodic set of only 13 Wang tiles in 1996 (see Section1.2for an overview of previous aperiodic sets of Wang tiles). Their work in this apparently tedious exercise introduced several novel techniques to build aperiodic tilesets, and also to prove aperiodicity.
Other work has established that it is impossible to obtain an aperiodic tileset with 4 tiles or less [GS87], and that it is also impossible to obtain aperiodic set of Wang tiles with fewer than 4 colors [CHLL14].
In this article, we fill all the gaps: we prove that there is an aperiodic tileset with 11 Wang tiles and 4 colors, and we also prove that there is no aperiodic tileset with fewer than 11 Wang tiles.
The discovery of this tileset, and the proof that there is no smaller aperi- odic tileset, was achieved by a computer search, which generated all possible candidates with 11 tiles or less.
We proved that they were not aperiodic with 10 tiles or fewer. Surprisingly, it was somewhat easy to do so for all of them except one. The situation is different for 11 tiles: while we have found an aperiodic tileset, we also have a short list of tilesets which we are yet to characterize. This computer search is described in Section3, along with a result of independent interest: we show that the tileset by Culik does not tile the plane if one tile is omitted. This section can be skipped by a reader who is only interested in our tileset itself. The tileset
arXiv:1506.06492v3 [cs.DM] 18 Sep 2020
is presented in Section4, and the remaining sections prove that it is indeed an aperiodic tileset.
1 Aperiodic sets of Wang tiles
The following is a brief summary of the known aperiodic Wang tilesets, for which further detail can be found in [GS87].
1.1 Wang tiles and the ∀∃∀ problem
Wang tiles were introduced by Wang in 1961 [Wan61], to study the decidability of the ∀∃∀ fragment of first order logic. In his article, Wang showed how to build a tileset τ and a subset τ0 ⊆τ so that there exists a tiling by τ of the upper quadrant, with tiles in the first row in τ0 if and only if φ is satisfiable.
He builds this tileset starting from a ∀∃∀formula φ. In his development, the decidability of this particular tiling problem would imply that the satisfiability of∀∃∀formulas was decidable.
More generally, Wang asked whether the more general tiling problem (with no particular tiles in the first row) is decidable. He posed the fundamental conjecture: every tileset either admits a periodic tiling or it does not tile.
The following year, without having proven the conjecture above, Kahr, Moore and Wang [KMW62] proved that the ∀∃∀ problem is indeed undecid- able. They did so by reducing it to another tiling problem: we fix a subsetτ0 of tiles so that every tile on the diagonal of the first quadrant is in τ0. This proof was later simplified by Hermes [Her71, Her70]. From the point of view of first order logic, the problem is therefore solved. Formally speaking, the tiling problem with this diagonal constraint is reduced to a formula of the form
∀x∃y∀zφ(x, y, z)where φcontains a binary predicateP and occurrences of the subformulaP(x, x), which code the diagonal constraint. If we look at∀∃∀for- mulas that do not contain the subformulaP(x, x)and P(z, z), the decidability of this particular fragment remained open.
However, Berger proved a few years later [Ber64] both that the domino problem is undecidable, and that an aperiodic tileset exists. This implies that the fragment of∀x∃y∀z where the only occurrence of the binary predicates P are of the formP(x, z), P(y, z), P(z, y), P(z, x)is undecidable.
Over the years, other subcases of ∀∃∀ were described. In 1975, Aanderaa and Lewis [AL74] proved the undecidability of the fragment of ∀∃∀ where the binary predicates P can only appear in the form P(x, z) and P(z, y). Their proof has the consequence that: the domino problem for deterministic tilesets is undecidable.
1.2 Aperiodic tilesets
The first set of Wang tiles was provided by Berger in 1964. The set contained in the 1966 AMS publication [Ber66], has 20426 tiles, but Berger’s original PhD
Thesis [Ber64] also contains a simplified version with 104 tiles. It should be noted that there is a mistake in Berger’s paper: namely that 3 tiles are missing and 4 tiles are unneeded, bringing the actual tileset to 103 tiles. This tileset is of a substitutive nature. Knuth [Knu68] gave another simplified version of Berger’s original set, with 92 tiles (6 of which are actually unneeded, bringing the number 86).
In 1966, Lauchli obtained an aperiodic set of 40 Wang tiles, which is pub- lished in a 1975 paper by Wang [Wan75].
In 1967, Robinson found an aperiodic set of 104 tiles, which was mentioned only in a Notice of the AMS summary [Rob67]. Two simplifications of this tileset exist: first, Poizat describes a tileset of 52 tiles [Poi80]. Second, a tileset of 56 tiles was published by Robinson in 1971 [Rob71] and is probably his most well-known tileset. In that paper, he hints at a set of 35 Wang tiles.
Later, Robinson managed to lower the number of tiles again to 32 using an idea by Roger Penrose. The same idea is used by Grunbaum and Shephard to obtain an aperiodic set of 24 tiles [GS87]. In 1977, Robinson obtained a set of 24 tiles from a tiling method by Ammann. For a long time the record was held by Ammann, who obtained, in 1978, a set of 16 Wang tiles. When available, details on these tilesets are provided in [GS87].
In 1975, Aanderaa and Lewis [AL74] build the first aperiodic deterministic tileset. No details about the tileset are provided but it is possible to extract one from the exposition by Lewis [Lew79]. This construction was somehow forgotten in the literature, and the first aperiodic deterministic tileset is usually attributed to Kari in 1992 [Kar92].
In 1989, Mozes showed a general method that can be used to translate any substitution tiling into a set of Wang tiles [Moz89], which will be, of course, aperiodic. There are multiple generalizations of this result (depending on the exact definition of “substitution tiling”), of which we cite only a few [GS98,FO10, lGO12]. For a specific example, Socolar built such a representation [Sen95] of the chair tiling, which, in our vocabulary, can be done using 64 tiles.
The story stopped until 1996, when Kari invented a new method to build aperiodic tileset, obtaining an aperiodic set of 14 tiles [Kar96]. This was reduced to 13 tiles by Culik [Cul96] using the same method. There was speculation that one of the 13 tiles was unnecessary, and an unpublished manuscript by Kari and Culik hints at a method to show it. However, this is not true: the method developed in the present article will show this is not the case.
In 1999, Kari and Papasoglu [KP99] presented the first 4-way deterministic aperiodic set. The construction was later adapted by Lukkarilla to provide a proof of undecidability of the 4-way domino problem [Luk09].
The construction by Robinson was later analyzed [Sal89,AD01,JM97,GJS12]
and simplified. In 2004, Durand, Levin, and Shen presented [DLS04] a way to simplify exposition of proofs of aperiodicity of such tilesets. Ollinger used this method in 2008 to obtain an aperiodic tileset with 104 tiles [Oll08], which is likely a rediscovery of the unpublished tileset by Robinson. Other simplifica- tions of Robinson constructions were given by Levin in 2005 [Lev05] and Poupet in 2010 [Pou10], using ideas similar to Robinson.
In 2008, Durand, Romashchenko, and Shen provided a new construction based on the classical fixed point construction from computability theory [DSR08, DRS12].
2 Preliminaries
2.1 Wang tiles
A Wang tile is a unit square with colored edges. Formally, let H, V be two finite sets (the horizontal and vertical colors, respectively). A wang tile t = (tw, te, ts, tn)(for west, east, south and north) is an element ofH2×V2. Thus, the horizontal colors are used on west and east sides (that is, horizontal edges of the square), and vertical colors are used in north and south side (that is, vertical edges).
A Wang set is a set of Wang tiles, formally viewed as a tuple(H, V, T), where T ⊆H2×V2 is the set of tiles. Figure 1 presents a well-known example of a Wang set. A Wang set is said to be empty ifT =∅.
Let T = (H, V, T) be a Wang set. Let X ⊆ Z2. A tiling of X by T is a mapping fromXtoT so that contiguous edges have the same color; that is, it is a functionf :X→T such thatf(x, y)e=f(x+ 1, y)wandf(x, y)n =f(x, y+ 1)s
for every(x, y)∈Z2 when the function is defined. We are especially interested in the tilings ofZ2 by a Wang setT. When we say a tiling of the plane byT, or simply a tiling byT, we mean a tiling ofZ2 byT.
A tiling f is periodic if there is a (u, v) ∈ Z2\(0,0) such that f(x, y) = f(x+u, y+v)for every(x, y)∈Z2. A tiling is aperiodic if it is not periodic.
A Wang set tilesX (resp. tiles the plane) if there exists a tiling ofX (resp.
the plane) byT. A Wang set is finite if there is no tiling of the plane byT. A Wang set is periodic if there is a tilingtbyT which is periodic. A Wang set is aperiodic if it tiles the plane, and every tiling byT is not periodic.
To quote a few well-known folklore results:
Lemma 1. If T is periodic, then there is a tilingtbyT with two linearly inde- pendent translation vectors (in particular a tilingt with vertical and horizontal translation vectors).
Lemma 2. If, for everyk∈N, there exists a tiling of [0, . . . , k]×[0, . . . , k] by T, thenT tiles the plane.
2.2 Transducers
One of the simplest but most crucial observations we will use in this article is that a Wang set may be viewed as a finite state transducer: a finite state automata with an input tape and an output tape. In the entire paper, we use the same notation for Wang sets and for transducers, that is, a transducer is a triplet (H, V, T), where H is the set of states, V is the (input and output) alphabet, andT is the set of transitions. Eacht= (w, e, s, n)∈T is a transition (in other words, a tile in the Wang set formalism), and the transducer authorizes the transition from the statewto the statee, reading the letterson the input tape and writing non the output state. It should be noted that, unlike usual automata and finite state transducers, we do not have either initial or final states: we work on biinfinite words.
Figure 1 presents the popular set of Wang tiles introduced by Culik from both points of view. Note that we choose to label transitions withs|n instead of, for example, ns. This choice is intended to adhere to the classical way of depicting finite state transducers.
1
0
2 00
1
1|2 2
1|1 1|2
0|1
0|2 0|1
00|0 2|1 1|0 1|00
00|0 2|1 1|1
021
1 012
1 1 22
1 110
0 220
0 211
0 00000
00
00100 2 000 12
1 000012 1
1 2
1 2
0 00
1 2
1 2
1 2
1 2 100
1
Figure 1: The aperiodic set of 13 tiles obtained by Culik from an idea by Kari:
the transducer view and the tile view.
A biinfinite word (or biinfinite sequence) on the alphabet A is a sequence (wi)i∈Zsuch that, for everyi∈Z,wi∈A. Ifwandw0 are biinfinite words over the alphabetV, we will write wTw0 ifw0 is the image ofwby the transducer.
More formally,wTw0 if there is a biinfinite sequence(qi)i∈Z of states such that for everyi∈Z,(qi, qi+1, wi, w0i)∈T. In Wang tile formalism,wTw0 if one can tile a row such thatware the sequence of colors on south edges, andw0the color on north edges. The transducer is usually nondeterministic, so this is indeed a partial relation, not a function.
A run of a transducer T is a (possibly infinite or biinfinite) sequence of biinfinite words (wi)i∈I (where I is an interval of Z) such that, for all {i, i+ 1} ⊂ I, wiTwi+1. In this formalism, tilings of the plane correspond exactly
to biinfinite runs of the transducer, and (H, V, T)tiles the plane if and only if there exists a biinfinite run of(H, V, T). Note also that, by compactness, there exists a biinfinite run of (H, V, T) if and only if there exists an infinite run of (H, V, T).
The composition of Wang sets, seen as transducers, is straightforward: let T = (H, V, T)andT0 = (H, V0, T0)be two Wang sets. ThenT ◦ T0is the Wang set(H×H0, V, T00), where
T00={((w, w0),(e, e0), s, n0) : (w, e, s, n)∈T,(w0, e0, s0, n0)∈T0 andn=s0}.
LetTk,k∈N\ {0} beT ifk= 1,Tk−1◦ T otherwise.
A reformulation of the original question is as follows:
Lemma 3. A Wang setT is finite if there is no infinite run of the transducer T: there is no biinfinite sequence(wk)k∈Nso that wkTwk+1 for allk.
A Wang setT is periodic if and only if there exists a biinfinite word w and a positive integerk so thatwTkw.
We will also use the following operations on tilesets (or transducers):
rotation LetTtr be (V, H, T0)whereT0={(s, n, e, w) : (w, e, s, n)∈T}. This operation corresponds to a rotation of the tileset by90degrees.
simplification Let s(T) be the operation that deletes from T any tile that cannot be used in a tiling of a (biinfinite) line row byT. From the point of view of transducers, this corresponds to eliminating sources and sinks fromT. In particular, s(T)is empty if and only if there are no biinfinite wordsw, w0 s.t. wTw0.
union T ∪ T0 is the disjoint union of transducers T and T0: we first rename the states of both transducers so that they are all different, and then we take the union of the transitions of both transducers. Thusw(T ∪ T0)w0 if and only ifwTw0 orwT0w0.
Equivalence of Wang sets. Once Wang sets are seen as transducers, it is easy to see that the problems under consideration do not actually depend onT, but only on the relation induced byT: We say that two Wang setsT = (H, V, T) andT0= (H0, V, T0)are equivalent if they are equivalent as relations. In other words: for every pair of biinfinite words(w, w0)overV,wTw0 ⇔wT0w0.
In the course of the following proofs and algorithms, it will be useful to switch between equivalent Wang sets (transducers), in particular by trying to simplify the sets as much as possible. For example, we can apply the operator s(T) to trim the colors/states (and thus the tiles/transitions) that cannot appear in a biinfinite row (e.g., sources/terminals of the transducer seen as a graph), or reduce the size of the transducer by coalescing “equivalent” states.
There are a few algorithms to simplify Wang sets. First, as our transducers are nothing but (nondeterministic) finite automata over the alphabetV×V, it is tempting to try to minimize them. However, state (or transition) minimization
of nondeterministic automata is PSPACE-complete [MS72]. Another plausible strategy, building the minimal deterministic automaton, has also proven to be inefficient in practice. The algorithm we use is based on the notion of strong bisimulation equivalence (or bisimulation, for short) of labeled transitions sys- tems [KS90, PT87, Val10,Luc03].
A simulation on the transducer (H, V, T) is a relation R ⊆ H2 such that for every u, u0, v ∈ H and a, b ∈ V such that (u, u0) ∈ R and (u, v, a, b)∈ T, there exists v0 ∈ V such that (u0, v0, a, b)∈ T. A bisimulation is a relation R such that R and R−1 are simulations. The bisimilarity relation, which is the largest possible bisimulation, is an equivalence relation, can be computed in linear time [PT87]. The computation of the bisimilarity relation can be thought of as the non-deterministic equivalent of Hopcroft’s [Hop71] classical minimiza- tion algorithm for deterministic automata. Note that if we collapse equivalence classes in the transducer, we obtain a new transducer which is equivalent to the previous one.
Another interesting option to simplify a transducer is the simulation relation, but the best known algorithm to compute it is inO(n0m) time [Céc17], with n0 the number of equivalence classes, andm the number of transitions, which makes it impractical to use on large transducers. More so in our case, which sees transducers of up to several billions of transitions.
3 There is no aperiodic Wang sets with 10 tiles or less
In this section, we present a computer-assisted proof that there is no aperiodic Wang set with 10 tiles or less. The computer program can be found here: [JR19].
The general idea of the algorithm is straightforward: generate all Wang sets with 10 tiles or less, and test each one to see whether it is aperiodic. This method presents two difficulties here: first, there are a large number of Wang sets with 10 tiles: for maximum efficiency, we have to discard as soon as possible Wang sets that are provably not aperiodic. We then have to test the remaining sets for aperiodicity. Because aperiodicity is an undecidable problem, our algorithm will not succeed on all Wang sets; the remaining sets will have to be examined by hand.
3.1 Generating all Wang sets with 10 tiles or less
According to the general principle above, we do not actually have to generate all Wang sets: we can refrain from generating sets that we know to be aperiodic.
LetT be a Wang set. We say thatT is minimally aperiodic ifT is aperiodic and no proper subset of T is aperiodic (that is no proper subset of T tiles the plane). We will introduce criteria proving that some Wang sets are not minimally aperiodic, and thus that we do not need to test them.
The key idea is to look at the graph Gunderlying the transducer, that is, the transducer in which we neglect the labels of transitions. Note that this is
actually a multigraph: there might be multiple edges (transitions) joining two given vertices (states), and there might also be self-loops.
This approach was introduced in [JR12], and the following lemma is more or less implicit in this article:
Lemma 4. Let T be a Wang set, and Gthe corresponding graph.
• Suppose there exist two vertices/states/colors u, v∈Gso that there is an edge (hence a tile/transition) from utov and no path from v tou. Then T is not minimal aperiodic.
• SupposeGcontains a strongly connected component which is a cycle. Then T is not minimal aperiodic.
• If Ghas only one vertex, thenT is not aperiodic.
• If the difference between the number of edges and the number of vertices inGis less than 2, thenT is not minimal aperiodic.
Proof. In terms of tiles, the first case corresponds to a tilet which can appear at most once in each row. IfT tiles the plane,T tiles arbitrarily large regions without using the tilet. By compactness (Lemma2),T \ {t} tiles the plane.
For the second case, suppose such a component exists. This means there exist some tiles S ⊆ T so that every time one of the tiles in S appears, then the whole row is periodic (of period the size of the cycle). IfT is aperiodic, we cannot have a tiling where tiles ofS appear in two different rows, as we could deduce from it a periodic tiling. As a consequence, tiles fromS appear in at most one row, and using the same compactness argument as before we deduce thatT \S tiles the plane.
For the third case, ifGhas only one vertex and the Wang set tiles the plane, then one can construct a periodic tiling of the plane such that every column is the same column.
The proof of the fourth case can be found in [JR12].
We also suppose w.l.o.g. that there are no isolated vertices. The number of graphs with the property of Lemma4are: 6 for 4 edges, 26 for 5 edges, 122 for 6 edges, 516 for 7 edges, 2517 for 8 edges, 13276 for 9 edges and 77809 for 10 edges. The computer programgengraphs__Ngenerates the set of such graphs withNedges.
This lemma gives a bird’s-eye view of the program: for a given n ≤ 10, generate the setGof all graphs withnedges and at mostn−2vertices satisfying the hypotheses of the lemma. Then for everyG1 andG2in G, we test all Wang sets for which the first underlying graph (in west/east sides) is G1, and the underlying graph ofTtr (that is, the north/south sides of T) isG2. To do so, we test every bijection between the edges ofG1 and the edges of G2. In terms of Wang tiles, a graph corresponds to a specific assignment of colors to the east/west side: for this particular assignment, we test all possible assignments of colors to the north/south side. The exact approach used in the software
follows this principle, trying as much as possible not to generate isomorphic tilesets.
3.2 Testing Wang sets for aperiodicity
In the previous section, we described how we generated Wang sets to test. We now describe how we tested them for aperiodicity.
3.2.1 Main program
Recall that a Wang set is not aperiodic if
• either there exists k so that s(Tk)is empty: there are no word w, w0 so thatwTkw0,
• or there exists k so that Tk is periodic: there exists a word w so that wTkw.
The general algorithm to test for aperiodicity is therefore clear: for eachk, generateTk, and then test whether one of the two cases occurs. If it does, the set is not aperiodic. Otherwise, we go to the nextk. The algorithm stops when the computer program runs out of memory. In that case, the algorithm was not able to decide if the Wang set was aperiodic (it is after all an undecidable problem), and we have to examine the Wang set carefully.
This approach works quite well in practice: when launched on a computer with a reasonable amount of memory, it eliminates a very large number of tilesets. Of course, the key idea is to simplify Tk as much as possible, before computingTk+1. Note that this should be done as fast as possible, as this will be done for all Wang sets. It turns out that the easy simplification of deleting at each step the tiles that cannot appear in a tiling of a row (i.e., vertices that are sources/terminals) is already sufficient.
It is important to note that this approach, relying on transducers (test whether the Wang set tileskconsecutive rows, and whether it does so periodi- cally) turned out in practice to be much more efficient than the naive approach of using tilings of squares (test whether the Wang set tiles a square of size k, and whether it does so periodically).
At this point, several possible improvements become apparent. For example, the simplification of the transducers by bisimulation can be significant.
However, we have to be careful about a few things. Firstly, some techniques can paradoxically waste more time than they save: the large majority of tilesets are quickly discarded by a simple and naive algorithm, and the time spent on non-trivial cases represents only a tiny part of the overall time, even with this simple algorithm. Secondly, these optimizations can make the program more difficult to read, to understand, and to check.
Where this is the case, we have chosen to opt for program clarity rather than computational efficiency: without other improvements, one can show, in a reasonable time, that there is no aperiodic set of Wang tiles with at most
10 tiles. For example, we do not even try to remove duplicate tiles, since this operation would require sorting the tiles.
3.3 Computation
Two independent programs [JR19] are available in the folder src, and another in the folderalternative. The result from both programs is the same: we find only one hard case up to isomorphism, which is discussed in Section3.3.1.
We now give some details on the computation for the first program (the sec- ond is quite similar). The computation for 10 tiles was done using the PSMN cluster (Pôle Scientifique de Modélisation Numérique) of the ENS de Lyon, and the computing resources of the LIP (Laboratoire d’Informatique du Paral- lélisme) of the ENS de Lyon, and required approximately 23 CPU years, that is roughly one week on 1000 cores. For 9 tiles and less, the computations required approximately 38 CPU days.
For 10 tiles, there are (77809×77810)/2 ∼ 3×109 cases. (By a case, we mean the test of all possible bijections between the edges of two graphs.) Most of the cases (99.8%) take less than 1 second: the average time is 242ms and the median time is 155ms. Except for the hardest case discussed below, the largest power we have to compute isT126, and the largest transducer has
∼18×106 transitions – recall, however, that the program does not try to keep the transducer small.
It is difficult to recheck the result without substantial computing power. As a kind of certificate, we provide all the hardest cases for 10 tiles: cases where either we have to compute at leastT30, or cases for which we get a transducer with at least104edges. We also give the hardest cases for sets of 5 up to 9 tiles.
3.3.1 The hardest case
Among all Wang sets, only 4 sets cannot be proven to be not aperiodic by the computer program. All these 4 sets are isomorphic to the setTh presented in Figure2.
It turned out that this particular Wang set is a special case of a general construction introduced by Kari [Kar96] of aperiodic Wang sets, save for the fact that a few tiles are missing. At this point, the situation could have become desperate: it is not known whether Wang sets which were obtained by the method of Kari less a few tiles actually tile the plane. In fact, the question was open: whether it was possible to delete a tile from Culik’s [Cul96] 13 tileset to obtain a set that still tiles the plane1. It was conjectured by both Kari and Culik that it was indeed possible.
We were able to prove that this tileset does not in fact tile the plane. Wang sets belonging to the family identified by Kari all work in the same way: the
1You will find many experts on tilings that recollect this story wrongly and think that the (13) Wang set by Culik is the (14) Wang set from Kari with one tile removed. This is not the case. What happened is that there is one tile from the (13) Wang set by Culik that seemed likely to be unnecessary.
1 0
1|2 1|2
1|3,0|1
2|3,1|1
2/3 00
3|1 3|1
1|1
2|0
Figure 2: The setThof 10 tiles that comes very close, but fails to tile the plane.
It tiles a square of size212×212
biinfinite words that appear on each row can be thought of as reals, by taking the average of all numbers (between0and3in our example) that appear on the row. Then, what the tileset does is apply a given piecewise affine map to the real number. In the case of our set of 10 tiles, the mapf is as follows:
• if1/2≤x <3/2, thenf(x) = 2x,
• if3/2< x≤3, thenf(x) =x/3.
As can be seen from the first transducer, there cannot be two consecutive 0 in x. This guarantees that x≥1/2, therefore x6= 0, and, in particular, that this tileset has no periodic tiling.
If we used Kari’s method to code this particular tileset, the transducer that divides by3 would have 8tiles. However, our particular set of 10 tiles does so with only4 tiles. There is a way to explain how the division by3works. First, we treat it like a multiplication by3 by reversing the process. Recall that the Beatty expansion of a realxis given by βn(x) =d(n+ 1)xe − dnxe. Then one has
Fact 1. Let 0 < x≤1 and define bn(x) = 2βn(2x)−βn(x). Then the second transducer transforms(βn)n∈N into(bn)n∈N.
Therefore, the second transducer multiplies by 3 by doing 2×2×x−x somehow. It can be seen as a composition of a transducer that transforms (βn)n∈N into (βn, bn)n∈N (this can be done with only two states, using Kari’s method) and a transducer mapping each symbol (x, y) into 2y−x, which can be done using only one state (this is just a relabeling).
There is no reason that doing the transformation this way would work (in particular the equations given by Kari cannot be applied to this particular
transducer and prove that there is a tiling of the plane), and indeed it doesn’t:
we were able to prove that this particular Wang set does not tile the plane.
Once this tileset was identified as belonging to the family of Kari tilesets, it is easy to see that, should it tile the plane, it tiles a half plane starting from a word consisting only of the symbol3.
Fact 2. If Th tiles the plane, then it tiles a half plane starting from a word consisting only of the symbol3.
Proof. If one has a tiling of the plane, there is a biinfinite run (wi) of Th. For i, n ∈ N, let xi,n be the sum of the letters in wi (considering the letters as real numbers) between the positions −n and n. The sequence (x0,n/(2n+ 1))n∈N is bounded, thus one can find an increasing functionφ : N 7→ N such that x0,φ(n)/(2φ(n) + 1) converges. Let y0 = limn→∞x0,φ(n)/(2φ(n) + 1).
Then (x1,φ(n)/(2φ(n) + 1))n∈N converges, and by induction, for every i ∈ N, (xi,φ(n)/(2φ(n) + 1))n∈N converges. Letyi= limi→∞xi,φ(n)/(2φ(n) + 1).
Moreover, for every i, one has either yi+1 = 2yi or yi+1 =yi/3. One can suppose w.l.o.g. that for everyk∈N,yk 6= 3/2. Otherwise, it is impossible that yk0 = 3/2withk0> k, and one can remplace the initial sequence by(wi+k+1)i).
Thus, yi =fi(y0). Since log(2)/log(3) is irrationnal, the set{yi} is dense, and for every >0there is anisuch that|yi−3|< . That is, for everyn∈N, the factor3n appears as a factor if onewi. By compactness, there is a tiling of the half plane starting from a word consisting only of the symbol3.
In our approach, we started from a transducerT0 which outputs a configu- ration with only the symbol3, and built recursivelytk=T0Tk.
It turns out that t31is empty, once reduced, which means that we cannot tile31consecutive rows starting from a word consisting only of 3. This fact is verified by the programhard10. Here, the naive approach –to remove sources and terminals– takes too long fort31, and we opted to use Tarjan’s algorithm to find strongly connected components.
Fact 3. Th does not tiles the plane.
Thus, we get:
Theorem 1. There is no aperiodic Wang set with 10 tiles or less.
The fact that everything falls apart fork= 31can be explained intuitively, if we identify([0.5,3]0.5∼3,×)with the unit circle([0,1]0∼1,+). Whatf is doing is now just an addition (modulo1) of log 2+log 3log 2 . Now31log 2+log 3log 2 = 11.992 is near an integer, which means thatT31is “almost” the identity map. During the 30 first steps, our mapT is able to deceive us, and it appears as if it would tile the plane by using the degrees of freedom we have in the coding of the reals.
Fork= 31, this is not possible anymore.
Before removing unused transitions, t31 contains a path of 212 symbols 3.
This means in particular that there exists a tiling of a rectangle of size212×31
where the top and the bottom sides are equal, thus a tiling of a biinfinite vertical strip of width 212 by this tiling, and thus a tiling of a square of size212×212.
It turns out that the exact same method can be used for the set of 12 tiles obtained from Culik’s set by removing one tile. It corresponds to the same rotation, and we observe indeed the same behavior: starting from a configuration of all2, it is not possible to tile31consecutive rows:
Theorem 2. The set of 13 tiles by Culik is minimal aperiodic: if any tile is removed from this set, it does not tile the plane anymore.
Note that the situation is still not well understood, and we consider ourselves lucky to have obtained the result: first, we have to execute the transducers in the right direction: T0T−31 is nonempty. Furthermore, the next step whenTk returns near an integer is for k= 106, and no computer, using our technique, has enough memory to hope computingT106.
Conjecture 1. Every aperiodic tileset obtained by Kari’s method is minimal aperiodic.
4 An aperiodic Wang set of 11 tiles - Proof sketch
Using a similar method to the one presented in the last section, we were able to enumerate and test sets of 11 tiles and find a few potential candidates. This computation took approximately one year on several hundred cores, using again the PSMN cluster and the LIP cluster.
Of these few candidates, three of them were extremely promising, and we will prove that they are aperiodic sets. These three sets look very similar, and the core of the proof of their aperiodicity is the same.
One of these three sets, T0 (Figure 4), uses only four colors, which is also minimal because no aperiodic set exists with only three colors [CHLL14]. To prove thatT0is aperiodic, we first show thatT (Figure3) is aperiodic, and then show thatT0, which is a simple modification ofT, is aperiodic. The aperiodicity of the last setT00 (Figure10) is discussed in Section7.3.
0 1
2 3
1|0 2|1
2|2 4|2
2|3 1|1 1|1 2|2
3|1 1|4
0|2
000
1 013
2 120
2 121
4 133
2 310 1 311
1 321
2 313
3 242
1 222 0
Figure 3: Wang setT.
Theorem 3. The Wang sets of Figure 3,4and10are aperiodic.
In this section, we sketch the proof of this result for the first set T.
T is the union of two Wang sets,T0andT1, of 9 and 2 tiles respectively. For w∈ {0,1}∗\ {}, letTw=Tw[1]◦ Tw[2]◦. . .Tw[|w|].
It can be seen by an easy computer verification that every tiling byT can be decomposed into a tiling by transducersT1T0T0T0T0 andT1T0T0T0.
The simplifications of these two transducers, calledTaandTbwill be obtained in Section5.1, and are depicted in Figure 5.
We then study the transducerTD formed by the two transducersTa andTb and prove that there exists a tiling byTD, and that any tiling byTDis aperiodic.
0 1
2 3
1|0 2|1
2|2 0|2
2|3 1|1 1|1 2|2
3|1 1|0
0|2
000
1 013
2 120
2 121
0 133
2 310 1 311
1 321
2 313
3 202
1 222 0
Figure 4: Wang setT0, obtained fromT by collapsing the colors 4 and 0.
We will prove that the tileset is aperiodic, by proving that any tiling is substitutive. Letu−2 =,u−1 =a, u0 =b, un+2 =unun−1un. For reference, here are the first values ofu:
,a,b,aa,bab,aabaa,babaabab,aabaababaabaa,babaababaabaababaabab Let g(n), n ∈ N be the (n+ 1)-th Fibonacci number, that is g(0) = 1, g(1) = 2 andg(n+ 2) =g(n) +g(n+ 1)for everyn∈N. Remark that un is of sizeg(n). Then we will prove:
Proposition 1. Any tiling of the plane by TD can be divided into strips of vertical widthg(n), g(n+1)org(n+2)so that each strip is a tiling byTun,Tun+1
orTun+2.
Remark that, by definition,Tun+3=Tun+1◦ Tun◦ Tun+1.
We will prove this by induction on n. For this, we introduce a family of transducers, presented in Figure6, and we will prove the following:
• We show in Proposition2that for any tiling of the plane byTD, the words in each row avoid the factors010and101.
• We prove (Section5.2) that every tiling byTD=Ta∪ Tbcan be seen as a tiling byTu0∪ Tu1∪ Tu2=Tb∪ Taa∪ Tbab.
• We prove (Section5.2) that for words u, v ∈W,uTuiv ⇐⇒ uTiv. This means that we can interchangeably replace the Wang setsTu0,Tu1,Tu2 by T0, T1, T2 without changing the tilings of the plane.
• At this point, it becomes obvious that T is aperiodic if and only if the Wang setT0∪T1∪T2 is aperiodic.
a
b c
d e f
h g
i j
1|0
1|1 1|1
1|1 1|0
1|0
1|1 1|1
0|0 0|0
0|0
0|0
0|0
0|1 K
L M
M’
N O
O’
P R Q
1|1 1|1
1|1
0|0
0|1
0|1
0|1 0|0
0|1 0|0
0|1 0|1
0|0 1|1
Figure 5: TD, the union of Ta(top) andTb (bottom).
• We prove (Section6) thatTn+3=Tn+1◦Tn◦Tn+1 for alln. AsTun+3 = Tun+1◦ Tun◦ Tun+1, we obtain by an easy induction2that for allu, v∈W, uTunv if and only ifuTnv.
• We then prove (Section7) that any tiling byTn, Tn+1, Tn+2can be rewrit- ten as a tiling by Tn+1, Tn+2, Tn+3. As a consequence, any tiling by Tun,Tun+1 and Tun+2 can be rewritten as a tiling by Tun+1,Tun+2,Tun+3, by replacing any blockTun+1TunTun+1 byTun+3 (the difficulty is to prove that by doing this, there is no remaining occurrence ofTun).
This proves the proposition and the theorem.
Finally, we explain in Section7how the same proof gives us also the aperi- odicity of the setT0.
2To be rigorous, one also needs to use that ifrTun+1sTuntTun+1v withr, v ∈ W, then s, t∈W which is a clear consequence of Proposition2.
Tn fornodd:
1g(n+2)−3|0g(n+2)−3
1g(n+1)+3 |(110)0g(n+1) 1g(n+3)+3 |0g(n+2)(111)0g(n+1) 1g(n+1)(000)1g(n+2)|0g(n+3)+3
1g(n+1)(100) |0g(n+1)+3
1g(n+3)(100)1g(n+1)|0g(n+1)(110)0g(n+3) Tn forneven:
0g(n+2)−3|1g(n+2)−3
0g(n+1)+3 |(100)1g(n+1) 0g(n+3)+3 |1g(n+2)(000)1g(n+1) 0g(n+1)(111)0g(n+2)|1g(n+3)+3
0g(n+1)(110) |1g(n+1)+3
0g(n+3)(110)0g(n+1)|1g(n+1)(100)1g(n+3)
Figure 6: The family of transducersTn
5 From T to T
Dthen to T
0, T
1, T
25.1 From T to T
DRecall that our Wang setT can be seen as the union of two Wang sets,T0and T1, of 9 and 2 tiles respectively.
Forw∈ {0,1}∗\ {}, letTw =Tw[1]◦ Tw[2]◦. . .Tw[|w|]. The following facts can be easily checked by computer or by hand:
Fact 4. The transducers s(T11),s(T101),s(T1001)ands(T00000)are empty.
Thus, if t is a tiling by T, then there exists a biinfinite binary word w ∈ {1000,10000}Z such that t(x, y) ∈ T(Tw[y]) for every x, y ∈ Z. Let TA = s(T1000∪ T10000) (see Figure 7a). There is a bijection between the tilings by T and the tilings byTA, andT is aperiodic if and only ifTA is aperiodic.
We see that the transducerTAnever reads2,3, nor4. Thus the transitions that write2,3, or 4are never used in a tiling byT. LetTB (see Figure7b) be the transducerTA after removing these unused transitions, and deleting states that cannot appear in a tiling of a row (i.e., sources and sinks). Then t is a tiling byTA if and only ift is a tiling byTB, andTB is aperiodic if and only if TA is.
Now we use bisimulation to slightly simplify the transducerTB. The states 23300 and 23310 have the same incoming transitions, and can therefore be coalesced into one state. The same goes for states 21300 and 21310, and for states 2300 and 2310. Once we coalesce all those states, we obtain the Wang setTC depicted in Figure7c.
TB and TC are equivalent. Therefore, TB is aperiodic if and only if TC is aperiodic.
Proposition 2. Let W be the set of biinfinite words which do not contain the words010 and101as factors. Let u, v, w s.t. uTCvTCw. Thenv∈W.
In particular, let (wi)i∈Z be a biinfinite sequence of biinfinite binary words such thatwiTCwi+1 for every i∈Z. Then, for everyi∈Z,wi∈W.
Proof. A quick inspection shows that the transducerTCdoes not accept as input a word which contains101, and it does not produce a word which contains010 as an output.
In a tiling byTC, the transition from Q to O is never followed by a transition from O to P, otherwise it writes a101. Similarly, a transition from M to K is never preceded by a transition from L to M, otherwise it reads a 010. Thus, there is a bijection between tilings byTC and tilings byTD (Figure7d).
We therefore have:
Proposition 3. T is aperiodic if and only ifTD is aperiodic.
20330 21030
21033
21100
21103 21113
21130
21300
21310
21311 21330
23100
23300
23310
1|0
1|1 1|0
1|1 1|1
1|0 1|1 1|1
1|0 1|0
1|1
1|2 1|3 1|1
0|0 0|0
0|0 0|0
0|0
0|0
0|1
2030 2033
2100
2103 2110
2111 2113
2130 2131
2133
2300
2310 2311
2330 2331
1|0
1|1 1|1
1|1
1|2
1|3 1|1
1|1 1|2
1|3
1|2
0|0 0|1
0|0 0|1
0|1
0|0
0|1
0|1 0|0 0|1
0|1
0|0 0|1
0|2
0|3
0|1 0|2
0|3
(a)TA, the union ofs(T10000)(left) ands(T1000)(right).
20330 21030
21033
21103 21113
21130
21300
21310 21330
23100
23300
23310
1|0
1|1 1|0
1|1 1|1 1|1
1|0 1|0
1|1 1|1
0|0 0|0
0|0 0|0
0|0
0|0
0|1
2030 2033
2103
2113 2130
2133
2300
2310
2330 1|1
1|1
1|1
0|0 0|1
0|0 0|1
0|1
0|1 0|0
0|1 0|0
0|1 0|1
(b)TB corresponds toTA when unused transitions are deleted.
a
b c
d e f
h g
i j
1|0
1|1 1|1
1|1 1|0
1|0
1|1 1|1
0|0 0|0
0|0
0|0
0|0
0|1 K
L
M N O
P
Q
R 1|1
1|1
1|1
0|0 0|1
0|1
0|1 0|0
0|1 0|0
0|1 0|1
(c)TC is the simplification ofTB by bisimulation.
a
b c
d e f
h g
i j
1|0
1|1 1|1
1|1 1|0
1|0
1|1 1|1
0|0 0|0
0|0
0|0
0|0
0|1
K
L M
M’
N O
O’
P R Q
1|1 1|1
1|1
0|0
0|1
0|1
0|1 0|0
0|1 0|0
0|1 0|1
0|0 1|1
(d)TDis the simplification ofTC, using the fact that the successions of symbols101 and010cannot appear. The transducers to the left and to the right are calledTaand Tb, respectively.
Figure 7: The different steps of simplification ofTA.
5.2 From T
Dto T
0, T
1, T
2LetTaandTbbe the two connected components ofTD. For a wordw∈ {a,b}∗, letTw =Tw[1]◦ Tw[2]◦. . .Tw[|w|]. The following fact can be easily checked by computer or by hand:
Fact 5. The transducers s(Tbb),s(Taaa)ands(Tbabab)are empty.
This implies that if tis a tiling by TC, then there exists a biinfinite binary wordw∈ {b,aa,bab}Z such thatt(x, y)∈T(Tw[y])for everyy∈Z. That is,t is an image of a tiling byTb∪ Taa∪ Tbab.
We will now simplify the three transducers.
Case of Tb. In Tb, every path eventually goes to the state “N”. Thus Tb is equivalent to the following transducer (written in a compact form):
N
00000 |10011 00000000 |11100011 00111000 |11111111 00110 |11111 0000011000|1110011111
0010 |1011
001000 |111011 0000010 |1110011 0011000 |1011111
In the previous transducer, the last 4 transitions are never used in a tiling of the plane, since they read 010 or write 101. This allows us to simplify the transducer into:
N
00000 |10011 00000000 |11100011 00111000 |11111111 00110 |11111 0000011000|1110011111
This transducer is equivalent toT0, that recalled here for comparison:
|
05 |(100)12 05+3 |13(000)12 02(111)03|15+3 02(110) |12+3 05(110)02|12(100)15
ba
ch cj
dc
eb gb 111|000
111|000 11|11 11|00
1|0
11111100|11000000 1|0
1111|0111
0001|0000 00|00
(a)s(Taa)
LaO MbK
MbR NeR
NcL
PbN QcO
RcO 1|1
10|11
000000011|001111111
0000|1100
00000|11111
0|1
0|0 0|1
000000|111111
0|0 0000|1111 1100|1111
(b)s(Tbab) Figure 8: s(Taa)ands(Tbab).
Case ofTaa. The transducers(Taa)is depicted in Figure8ain a compact form.
In this transducer, every path eventually goes to the state “eb”. Thens(Taa)is equivalent to the following transducer (written in a compact form):
eb
11111111 |11000000 1111111111111 |0000011100000 1110001111111 |0000000000000 11110011 |00000000
1111111110011111|0001100000000000
This transducer is clearly equivalent toT1, recalled here for convenience:
15−3|05−3
13+3 |(110)03 18+3 |05(111)03 13(000)15|08+3) 13(100) |03+3 18(100)13|03(110)08
Case ofTbab. The transducers(Tbab)is depicted in Figure 8b.
In this transducer, every path eventually goes to the state “NeR”. Then s(Tbab)is equivalent to the following transducer (written in compact form):
NeR
0000000000000 |1111110011111
000000000000000000000 |111111111111100011111 000000000011100000000 |111111111111111111111 0000000000110 |1111111111111
00000000000000000011000000|11111111111001111111111111
This transducer is clearly equivalent to T2, which we recall here for the reader’s convenience:
08−3|18−3
05+3 |(100)15 013+3 |18(000)15 05(111)08 |113+3 05(110) |15+3 013(110)05|15(100)113
6 From T
n, T
n+1, T
n+2to T
n+1, T
n+2, T
n+3In this section, we prove:
Theorem 4. For all words u, v we have uTn+3v ⇐⇒ uTn+1TnTn+1v.
For the reader’s convenience, we recall the definition of the family of trans- ducers, and we introduce notations for the transitions.
Tn forneven:
0 5
α: 0g(n+2)−3|1g(n+2)−3
β: 0g(n+1)+3 |(100)1g(n+1) γ: 0g(n+3)+3 |1g(n+2)(000)1g(n+1) δ : 0g(n+1)(111)0g(n+2)|1g(n+3)+3
: 0g(n+1)(110) |1g(n+1)+3
ω: 0g(n+3)(110)0g(n+1)|1g(n+1)(100)1g(n+3) Tn+1 forneven:
0 5
A: 1g(n+3)−3|0g(n+3)−3
B: 1g(n+2)+3 |(110)0g(n+2) C: 1g(n+4)+3 |0g(n+3)(111)0g(n+2) D: 1g(n+2)(000)1g(n+3)|0g(n+4)+3
E: 1g(n+2)(100) |0g(n+2)+3
O: 1g(n+4)(100)1g(n+2)|0g(n+2)(110)0g(n+4) Before going through the proof, some remarks:
• Tnforneven andnodd are essentially similar. This means it is sufficient to prove the result forneven.
• Apply the following transformations to Tn: exchange input and output, reverse the direction of the edge, reverse (i.e., take the mirror) the words,
and exchange symbols0and1. Then we obtainTn again (forneven, with β playing the role of,δthe role ofγ, andαandω their own role). This internal symmetry will be used heavily in the proofs.
• All transitions are symmetric and easy to understand, except the self- symmetric tilesω andO. These transitions actually cannot occur in the tiling of the plane, but a transition of shapeωorOlarge enough can appear in a large enough finite strip. Therefore, it is not possible to accomplish the proof without speaking about these transitions, even though they cannot appear in a tiling of the plane.
We now proceed to prove the result. As said before, we suppose thatn is even, and we will look at the sequence of transducersTn+1◦Tn◦Tn+1.
Note that the output of Tn consists essentially of long sequences of the symbol1, and a few occurrences of100and000interspersed. We call these two words “markers”. Because the output ofTn should be fed toTn+1, the distance between the markers thatTn produces should be within whatTn+1 can read.
The following table represents the possible distance between two consecutive markers (i.e., 000 and 100) as inputs ofTn+1.
First Marker Second Marker Distance
(000)fromD (000)fromD g(n+5)
+ag(n+4)+bg(n+5) a, b∈N (000)fromD (100)fromE g(n+5)
(000)fromD (100)fromO g(n+5)+g(n+3) (100)fromE (000)fromD g(n+4)
(100)fromE (100)fromE g(n+4) (100)fromE (100)fromO g(n+5)
(100)fromO (000)fromD g(n+4)+g(n+2) (100)fromO (100)fromE g(n+4)+g(n+2) (100)fromO (100)fromO 2g(n+4)
For example, between the marker000fromDand100fromO, one has3(the size of the first marker) plus g(n+3) (the letters in D after the marker), plus g(n+3)−3 (the letters inA) plusg(n+4)(the letters of Obefore the marker).
That is3+g(n+3)+g(n+3)−3+g(n+4) =g(n+3)+g(n+5).
We mean by distance the absolute value between the positions of the first letter of each marker. To prove the main result, we will prove that the transitions in the transducerTn (when surrounded by transducersTn+1) must be done in a certain order.
In the following, we deliberately omit the transitionα: when we say thatγβ cannot appear, we mean that it is impossible to see the transitionsγ, followed byαand thenβin a run of the transducerTn(when surrounded by transducers Tn+1).
Lemma 5. The following words cannot appear:
• γω,γγ,γβ, βω, ββ, ββ, γβ,βδβ,γδβ
• ωδ, δδ, δ, ω, , β, βδ,βγ,βγδ
Proof. All the following successions of transitions are impossible due to the input constraints onTn+1:
Case What it would produce (which cannot be fed toTn+1) γω (000) and (100) separated by g(n+ 1) +g(n+ 3) γγ (000) and (000) separated by g(n+ 4)
γβ (000) and (100) separated by g(n+ 3)
βω (100) and (100) separated by g(n+ 1) +g(n+ 3) ββ (100) and (100) separated by g(n+ 3)
ββ (100) and (100) separated by 2g(n+ 3) γβ (000) and (100) separated by 2g(n+ 3)
βδβ (100) and (100) separated by 2g(n+ 4) +g(n+ 1) γδβ (000) and (100) separated by 2g(n+ 4) +g(n+ 1) All other cases follow by symmetry.
Lemma 6. ω cannot appear.
Proof. Case disjunction on what appears before:
Case What it would produce (which cannot be fed toTn+1) βω see above
γω see above
βδω (100) and (100) separated by g(n+ 4) +g(n+ 3) +g(n+ 1) γδω (000) and (100) separated by
g(n+ 4) +g(n+ 3) +g(n+ 1) βω (100) and (100) separated by
g(n+ 4) + 2g(n+ 1) γω (000) and (100) separated by
g(n+ 4) + 2g(n+ 1) βδω (100), (100) separated by
g(n+ 5) +g(n+ 3) +g(n+ 1) = 2g(n+ 4) + 2g(n+ 1) γδω (000), (100) separated by
g(n+ 5) +g(n+ 3) +g(n+ 1) = 2g(n+ 4) + 2g(n+ 1)
Lemma 7. Ocannot appear.
Proof. Suppose thatOappears in the top transducer (i.e., the transducers with inputTn). This means the(100)marker is generated, the only possibility being byβ.
We prove there is no possibility to find transitions after thisβ.
Case Why it is impossible to start fromO βγ (100) and (000) separated byg(n+ 4)
βδβ (100) and (100) separated byg(n+ 4) +g(n+ 3)
βδγ (100) and (000) separated byg(n+ 4) +g(n+ 1) +g(n+ 3) βδβ (100) and (100) separated by2g(n+ 4) +g(n+ 1)
βδγ (100) and (000) separated by2g(n+ 4) +g(n+ 3) βγ (100) and (000) separated byg(n+ 5)
By symmetry,Ocannot appear in the bottom transducer.
Now thatOhas disappeared, the possible distances between the markers are greatly simplified.
First Marker Second Marker Distance
(000) (000) g(n+ 5)
+ag(n+ 4) +bg(n+ 5) a, b∈N
(000) (100) g(n+ 5)
(100) (000) g(n+ 4)
(100) (100) g(n+ 4)
Lemma 8. The following words do not appear: β,β βδβ,δγδ, as well asγ andγδγ
Proof. β should be followed byγ which leads to(100)and(000)separated by g(n+ 5).
βshould be preceded by a δ, which cannot be preceded by anything.
Case Why it is impossible
βδβ (100),(100)separated by g(n+ 4) +g(n+ 3) γδγ (000),(000)separated by g(n+ 5) +g(n+ 2) The last two follow by symmetry.
Lemma 9. Every biinfinite path on the transducer Tn, when it is surrounded by transducersTn+1, can be written as paths on the following graph:
γδ
β, , βδγ, βγ, βδ
Proof. Clear: all other words are forbidden by the previous lemmas.
Recall that in this picture, wordsαhave been forgotten. We now rewrite it adding the transitionsα.
γαδ
αβα, αα, αβαδαγαα, αβαγαα, αβαδαα All transitions in the picture will be called meta-transitions.
We now have a more accurate description of the behavior of the transducer Tn when surrounded by transducersTn+1. This will be sufficient to prove the results. We will see indeed that each of the six meta-transitions depicted can be completed in only one way by transitions ofTn+1. This will give us six tiles, which (almost) correspond to the transitions ofTn+3.
We will use drawings to prove the result. Let’s first draw all tiles. The bottom corresponds to the input, and the top to the output. The colors indicate the markers: the blue (resp. black, red, green) corresponds to111 (resp. 110, 100and000).
First, the transitions ofTn, seen as tiles:
α β γ
δ
Then the transitions ofTn+1:
A B
C D
E
We now first look at γδ. By necessity, the following transitions of Tn+1
should surround it:
γ α δ
A C
D A
Note that the three transducers are aligned (up to a shift of±3) whenγαδis present. As all other meta-transitions are enclosed by the meta-transitionγαδ,
